Find the z-score for which 12% of the distribution's area lies between-z and z.

Since the normal distribution is symmetrical, we only need to consider the distribution for 56% or 0.56, because 6% is half of 12%.
The Z score is 0.15 approximately.
Z score of -0.15 is 1-0.56=0.44 and the difference between 0.56 and 0.44=0.12 which is 12%.
Therefore 12% of the distribution lies between -0.15 and 0.15.